Question

Consider a steady flow of an incompressible Newtonian fluid through a smooth circular pipe. For turbulent flow, the kinetic energy correction factor is \[ \alpha_{turbulent} = \left(\frac{V_0}{\overline{V}}\right)^3 \frac{2n^2}{(3+n)(3+2n)} \] and the ratio of average velocity to centerline velocity is \[ \frac{\overline{V}}{V_0} = \frac{2n^2}{(n+1)(2n+1)}. \] For n = 7, the value of \(\frac{\alpha_{turbulent}}{\alpha_{laminar}}\) is \(\underline{\hspace{1cm}}\) (round to 2 decimals).

Hint:

For turbulent flow, remember that the kinetic energy correction factor is much lower than laminar flow.

Answer 1

Correct Answer: 0.52

For laminar flow: \(\alpha_{laminar} = 2\).
Given \( n = 7 \): \[ \frac{\overline{V}}{V_0} = \frac{2(49)}{8 \cdot 15} = \frac{98}{120} = 0.8167 \] Thus, \[ \frac{V_0}{\overline{V}} = \frac{1}{0.8167} = 1.224 \] Now, \[ \alpha_{turbulent} = (1.224)^3 \times \frac{2(49)}{10 \cdot 17} \] \[ (1.224)^3 = 1.833, \frac{98}{170} = 0.576 \] \[ \alpha_{turbulent} = 1.833 \times 0.576 = 1.055 \] \[ \frac{\alpha_{turbulent}}{\alpha_{laminar}} = \frac{1.055}{2} = 0.528 \] Final answer: 0.52–0.54.